Jump to content


Welcome to BrainDen.com - Brain Teasers Forum

Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account.
As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends.

Of course, you can also enjoy our collection of amazing optical illusions and cool math games.

If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top.
If you have a website, we would appreciate a little link to BrainDen.

Thanks and enjoy the Den :-)
Guest Message by DevFuse
 

Photo
- - - - -


  • Please log in to reply
50 replies to this topic

#21 adiace

adiace

    Advanced Member

  • Members
  • PipPipPip
  • 113 posts

Posted 01 June 2009 - 10:29 PM

My answer to the second question is wrong.. well incomplete. It is true only if A is successful

Edited by adiace, 01 June 2009 - 10:30 PM.

  • 0

#22 palmerc7

palmerc7

    Senior Member

  • Members
  • PipPipPipPip
  • 532 posts

Posted 01 June 2009 - 10:32 PM

If you look at the 1st table i posted...

If A succeeds at killing some one, then B must shoot A. Say I am right and A shoots D and kills him, now B can shoot A, C, or E. If B soots A and succeed, then C will have to shoot B, D or E. If C kills E, then D will kill C so that he can face the less successful B-bullet. If C kills D, then E will kill C (he too would rather have B shoot that him than C). So C must kill B. To prevent C from killing him, B must kill A.


That makes no sense to me. If A kills someone, then his chances of killing next time are even lower, theoretically. His chances are much lower of killing 2 in a row. So he would be the absolute least threatening, besides the fact that A has already went and it makes more sense to try and take out someone yet to go.
  • 0

#23 James8421

James8421

    Advanced Member

  • Members
  • PipPipPip
  • 468 posts

Posted 01 June 2009 - 10:39 PM

If you look at the 1st table i posted...

If A succeeds at killing some one, then B must shoot A. Say I am right and A shoots D and kills him, now B can shoot A, C, or E. If B soots A and succeed, then C will have to shoot B, D or E. If C kills E, then D will kill C so that he will have to face D-bullet rather than the less successful B-bullet. If C kills D, then E will kill C (he too would rather have B shoot that him than C). So C must kill B. To prevent C from killing him, B must kill A.



I don't get why B has to shoot A if A hits someone. I would think to myself, 'ok A got lucky and hit E or D, but his chances of hitting again are still 20%,' So I'd still go after D or E since they still have the higher chance of hitting.

If A,B,and C; between the three of them, could manage to eliminate D and E, Then C has the upper hand in the next rounds. But in reality A and B will probably miss, so C should shoot B, and let D and E shoot at each other. Then A and C try to eliminate E or D. C has a good chance at hitting. So that leaves A and C, with C being the likely victor. So I say C has best chance of surviving. Also thats what I would pick to be when the warden asks.
  • 0

#24 palmerc7

palmerc7

    Senior Member

  • Members
  • PipPipPipPip
  • 532 posts

Posted 01 June 2009 - 10:48 PM

I don't get why B has to shoot A if A hits someone. I would think to myself, 'ok A got lucky and hit E or D, but his chances of hitting again are still 20%,' So I'd still go after D or E since they still have the higher chance of hitting.

If A,B,and C; between the three of them, could manage to eliminate D and E, Then C has the upper hand in the next rounds. But in reality A and B will probably miss, so C should shoot B, and let D and E shoot at each other. Then A and C try to eliminate E or D. C has a good chance at hitting. So that leaves A and C, with C being the likely victor. So I say C has best chance of surviving. Also thats what I would pick to be when the warden asks.


That still doesn't make too much sense, or at least choosing C if you had the choice. D has a 4/5 chance of taking out E and if he doesn't, D is gone. So one of them is guaranteed to make it out of the first round, if you guess that ABC go after each other. second round, whichever is left is gunning for C now. Your logic assumes a kill, which in the case of A and B, very unlikely.
  • 0

#25 James8421

James8421

    Advanced Member

  • Members
  • PipPipPip
  • 468 posts

Posted 01 June 2009 - 11:16 PM

That still doesn't make too much sense, or at least choosing C if you had the choice. D has a 4/5 chance of taking out E and if he doesn't, D is gone. So one of them is guaranteed to make it out of the first round, if you guess that ABC go after each other. second round, whichever is left is gunning for C now. Your logic assumes a kill, which in the case of A and B, very unlikely.


Anyway you test it out you have to assume someone other than E will get a kill. I'm just assuming C kills B in first round. and either D hits E or E hits D if D misses, either way the order starts over leaving A,C, and D or E. Now D or E will still be targeted because they have better chances of kill. So A tries, and fails. Now C goes and say he hits. That leaves A and C, with C having the best chance of surviving.

Basically the way I see it is the two guys with the worst guns (A,B) will try for the two guys with the best gun (D,E). The guys with the best gun will try to eliminate each other, and the order starts over, and C has yet to be targeted. That leaves at the most one guy with a good gun (DorE). So A B and C target him, before he gets a chance at shooting. Then that leaves A B C. C still has a better chance of surviving.

E will be targeted the most, so his chance of survival is slim.
D will be targeted the second most, once again slim chance.
C will be targeted 3rd most, but by the 2 worst guns, and if it came down to it I'd rather be C, and hope B misses, and blast B. And have A and C left in the final round where C has a better chance.

So I'll still stick with C. :thumbsup:
  • 0

#26 final

final

    Advanced Member

  • Members
  • PipPipPip
  • 271 posts

Posted 01 June 2009 - 11:37 PM

yah im pretty sure you wanna choose a,b,or c and im currently thinking b beats out A by a smidge. Logic in previous post but to reiterate a lil of it. but now c logic is simply in first round d or e are gone (most likely only one but not definely and very unlikely more then those two). Then you either have d or e shoot at c (c is prolly dead) or a and b shooting at c (decent chance at survival) but thats a decent chance of survival in a unlikely (not too unlikely) scenario. Where in my previous logic A will be in the last two (most likely with b or c) and a wins something like ... well logic is in previouos post but i say B is the winner closely followed by a. I used alot of close-ities in my math but im pretty sure that i applied them evenly enough across the board that the ratio of the percentages is pretty close to the ratios of the real ones
  • 0

#27 adiace

adiace

    Advanced Member

  • Members
  • PipPipPip
  • 113 posts

Posted 01 June 2009 - 11:53 PM

Spoiler for and my final ans is..



Spoiler for Explanation

  • 0

#28 adiace

adiace

    Advanced Member

  • Members
  • PipPipPip
  • 113 posts

Posted 02 June 2009 - 12:12 AM

I am am not very clear at times....

This is why B must kill A if A succeeds at killing C, D or E

Say A kills C.
Then B has to kill A, D or E.

> If B kills D, then E will kill B, and be shoot at by A on the next turn (getting shot at by A is better for E)
> If B kills E, then D will kill B, and be shot at by A on the next trun.
> If B kills A, then D will kill E ( or E will kill D on his turn), and then it will be B's turn, and he gets to shoot at D.

Of the three cases, B killing A gives him the best chance of survival.

Edited by adiace, 02 June 2009 - 12:15 AM.

  • 0

#29 adiace

adiace

    Advanced Member

  • Members
  • PipPipPip
  • 113 posts

Posted 02 June 2009 - 12:43 AM

The bonus question... the order is still A to E right?

Edited by adiace, 02 June 2009 - 12:46 AM.

  • 0

#30 final

final

    Advanced Member

  • Members
  • PipPipPip
  • 271 posts

Posted 02 June 2009 - 01:33 AM

so i see my mistake if a person has two people above him he doesnt want to take a shot at the highest anymore (a person never wants to chance leaving only one person above him because this means the best shot left in the game will shoot at you) but that is the only reason i see. You want to take out the highest person there is without it making you the next target so i say. A shoots at E. B shoots at E if A missed otherwise he only has two above him so he shoots at A. C shoots at D if E is dead and B (the highest that doesnt make him an immediate target) otherwise. D shoots at E if alive or C if E isnt alive. E shoots at D if e is alive. so D has the best chance for the first round.

ok the reason a shoots a E not C. Think of this as you have your place and people above and below. The only thing that matters to you is that you have at least two people above you or none so the people above you can fight it out or there are none. Now all the people above you are just identical items. The only thing that matters is if you want to kill someone above you or below you. If you kill c the logic for your opponents d and e is the exact same as if you kill e for c and d. They also only care how many are above and below them. So if you(b at the time) kill one aboive you then have one of the people above you has one up two down and one has three down. Thats what determines there new logic. There old spot is mute. So since it doesnt matter who you kill for there logic you might as well take out the best shot of the equals. So if your aiming below yourself aim one below. If your aiming above aim all the way.

now i need to rework the second part

Edited by final, 02 June 2009 - 01:34 AM.

  • 0




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users